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Câu hỏi: Cho $\int{\dfrac{\cos 2x}{{{(\sin x+\cos x+2)}^{3}}}dx}=-\dfrac{{{(\sin x+\cos x+1)}^{n}}}{{{(\sin x+\cos x+2)}^{n}}}+C$ với $m,n\in \mathbb{N}$. Tính $A=2m+3n$.
A. $A=8$
B. $A=10$
C. $A=9$
D. $A=7$
A. $A=8$
B. $A=10$
C. $A=9$
D. $A=7$
$I=\int{\dfrac{\cos 2x}{{{(\sin x+\cos x+2)}^{3}}}dx}=\int{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{(\sin x+\cos x+2)}^{3}}}dx}=\int{\dfrac{(\cos x+\sin x)(\cos x-\sin x)}{{{(\sin x+\cos x+2)}^{3}}}dx}$
Đặt $t=\sin x+\cos x+2\Rightarrow dt=(\cos x-\sin x)dx$
$\Rightarrow I=\int{\dfrac{t-2}{{{t}^{3}}}dt}=\int{\left( \dfrac{1}{{{t}^{2}}}-\dfrac{2}{{{t}^{3}}} \right)dt}=-\dfrac{1}{t}+\dfrac{1}{{{t}^{2}}}+C=\dfrac{1-t}{{{t}^{2}}}+C=-\dfrac{\sin x+\cos x+1}{{{(\sin x+\cos x+2)}^{2}}}+C$
$\Rightarrow m=1;n=2\Rightarrow A=2.1+3.2=8$.
Đặt $t=\sin x+\cos x+2\Rightarrow dt=(\cos x-\sin x)dx$
$\Rightarrow I=\int{\dfrac{t-2}{{{t}^{3}}}dt}=\int{\left( \dfrac{1}{{{t}^{2}}}-\dfrac{2}{{{t}^{3}}} \right)dt}=-\dfrac{1}{t}+\dfrac{1}{{{t}^{2}}}+C=\dfrac{1-t}{{{t}^{2}}}+C=-\dfrac{\sin x+\cos x+1}{{{(\sin x+\cos x+2)}^{2}}}+C$
$\Rightarrow m=1;n=2\Rightarrow A=2.1+3.2=8$.
Đáp án A.